Решите уравнение номер 8 заменой переменной

[latex]\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\; \; ,\; \; ODZ:\; \left \{ {{3x^2+5x+8 \geq 0} \atop {3x^2+5x+1 \geq 0}} \right. \\\\t=3x^2+5x+1\; ,\; \; \; \sqrt{t+7}-\sqrt{t}=1\\\\\sqrt{t+7}=\sqrt{t}+1\\\\t+7=t+2\sqrt{t}+1\\\\2\sqrt{t}=6\; \; ,\; \; \sqrt{t}=3\; \; ,\; \; t=9\\\\3x^2+5x+1=9\\\\3x^2+5x-8=0\\\\D=25+4\cdot 3\cdot 8=121\\\\x_1=\frac{-5-11}{6}=-\frac{8}{3}=-2\frac{2}{3}\; ,\; \; x_2=\frac{-5+11}{6}=1\\\\Proverka:\; \; a)\; \; x=-\frac{8}{3}\; :\\\\\sqrt{ \frac{64}{3} - \frac{40}{3} +8}-\sqrt{ \frac{64}{3}-\frac{40}{3}+1}=[/latex][latex]=\sqrt{ \frac{48}{3} }-\sqrt{\frac{27}{3}}=\sqrt{16}-\sqrt9}=4-3=1\\\\1=1\\\\b)\; \; x=1:\\\\\sqrt{3+5+8}-\sqrt{3+5+1}=\sqrt{16}-\sqrt{9}=4-3=1\\\\1=1\\\\Otvet:\; \; x=-\frac{8}{3}\; ,\; \; x=1\; .[/latex]